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SubjectsChemistry
असामान्य आणविक द्रव्यमान
In Class 12 Chemistry, the topic Abnormal Molecular Mass explains why the molar mass calculated from colligative properties may differ from the actual molecular mass. Students learn how solute particles associate, such as through dimerisation, or dissociate into ions in solution, changing the number of particles present. They use the van’t Hoff factor to relate these changes to relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure, and calculate the degree of association or dissociation in solutions.
TOPIC PRACTICE
Up to 4 questions from this page. Select your focus, then start.
Take the initial amount of solute as one mole. The 60% that remains normal contributes 0.60 mole of particles. The remaining 40% forms dimers, and every dimer contains two original molecules, so this part contributes 0.40/2 = 0.20 mole of dimers. The total particle amount is therefore 0.60 + 0.20 = 0.80 mole, giving i = 0.80.
One formula unit of A₂B₃ gives five ions on complete dissociation: 2A and 3B. For 60% dissociation, the van’t Hoff factor is i = 1 + α(n − 1) = 1 + 0.60(5 − 1) = 3.4. Effective moles of particles are the initial moles multiplied by i: 0.01 × 3.4 = 0.034 mol. Thus option C is correct.
AB₃ produces four ions on complete dissociation, so for 25% dissociation the van’t Hoff factor is i = 1 + 0.25(4 − 1) = 1.75 = 7/4. For dissociation, the observed molar mass is related to the true molar mass by M_obs = M_true/i. Hence M_obs/M_true = 1/(7/4) = 4/7. The observed value is smaller because dissociation increases the number of particles.
A 1:3 electrolyte produces 1 + 3 = 4 ions when completely dissociated. For partial dissociation, use i = 1 + α(n − 1), where α = 0.50 and n = 4. Thus i = 1 + 0.50(4 − 1) = 1 + 1.5 = 2.5. The value 4.0 would apply only to complete dissociation, while 1.5 does not account for all three additional ions.
QUIZ COMPLETE
